Calculating Work for a Constant Velocity Elevator

AI Thread Summary
The discussion centers on calculating the work done by the tension in a cable lifting a 1200 kg elevator at a constant velocity over 35 meters. Initially, there was confusion regarding the work being zero due to constant velocity implying zero acceleration and force. However, it was clarified that while the net force is zero, the work done by the tension is positive, and the work done by the elevator's weight is negative, resulting in equal magnitudes of work. The correct answers are 4.1 x 10^5 J for the tension and -4.1 x 10^5 J for the weight. This highlights the distinction between net work and the work done by individual forces.
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Homework Statement



A cable lifts a 1200
kg elevator at a constant velocity for
a distance of 35 m. What is the work done by
(a)
the tension in the cable and
(b)
the elevator's weight?


Homework Equations



W = fscos(theta)

The Attempt at a Solution



okay i already know the answer to the solution and how to do it, however, when i first interpreted the question i thought the answer was zero because it says that it moves at a constant velocity. if its at a constant velocity then acceleration is zero. since f = ma, and acceleration is zero then force is equal to zero, so that means work is also equal to zero since W=fscos(theta)
however the answer to the question is actually

Solution: (a) 4.1 x 10 5J, (b) -4.1 x 10 5J

again, i know how to get the right answer but i would like to know what was wrong with my initial thinking
 
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hey123a said:

Homework Statement



A cable lifts a 1200
kg elevator at a constant velocity for
a distance of 35 m. What is the work done by
(a)
the tension in the cable and
(b)
the elevator's weight?


Homework Equations



W = fscos(theta)

The Attempt at a Solution



okay i already know the answer to the solution and how to do it, however, when i first interpreted the question i thought the answer was zero because it says that it moves at a constant velocity. if its at a constant velocity then acceleration is zero. since f = ma, and acceleration is zero then force is equal to zero, so that means work is also equal to zero since W=fscos(theta)
however the answer to the question is actually

Solution: (a) 4.1 x 10 5J, (b) -4.1 x 10 5J

again, i know how to get the right answer but i would like to know what was wrong with my initial thinking

The net force is zero.Hence net work done is zero.But the work done by the tension in the cable and the weight are non zero.

You can easily deduce this from the answers you have got.The magnitude of the work done by the two forces are equal,but that of the tension is positive while that of the weight is negative.
 
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